Torque on a dam

1. Sep 2, 2015

bigplanet401

1. The problem statement, all variables and given/known data
A body of water of depth D sits behind a vertical dam. The water and dam are in static equilibrium. Calculate the torque on the dam due to the water about an axis at ground level (that is, a depth D below the surface of the water).

2. Relevant equations
N (torque) = r x F

3. The attempt at a solution

I'm guessing that the force on the dam due to the water acts at a point D/2 above ground since the vertical coordinate of the center of mass of the water would be this high. Then the torque is just DF/2.

But consider the following argument. The force on the dam varies linearly with depth. Wouldn't this mean that I would get the same answer if I divided the water into N "slabs" (like a stack of books) and calculated the contribution to the torque from each slab, each with its own moment arm?

I tried the following: divide the water into N "slabs" of thickness D/N. The height of the ith slab is Di/N. The force on the dam from this slab is F(1 - i/N), where the maximum force (at the bottom of the dam) is F. Then the total torque is

$$\sum_{i=1}^N \; \frac{FDi}{N} \left( 1 - \frac{i}{N} \right)$$

But if I try to work out the sum and take the limit as N -> infinity, the sum becomes infinite. Is it just my math, or is this not the right argument? Thanks!

2. Sep 2, 2015

TSny

As you suggest, your answer of DF/2 is not correct.

For your summation approach, note that the force on a slab is due to fluid pressure acting on the slab. How does the force on a slab depend on the area of the slab? What happens to the areas of the slabs as N increases? You need to take that into account.

3. Sep 3, 2015

bigplanet401

Thanks for the hints. After thinking about it more, I came up with the following sum for the torque:

$$N = \sum_{i=1}^N \; \left(D - \frac{Di}{N} \right) \rho W g \frac{D}{N} \frac{Di}{N}$$

where D is the depth of the water, N is the number of slabs (each of thickness D/N), W is the width of the dam, rho is the density of water, and g is the acceleration due to gravity. I did the sum, took the large-N limit, and got

$$N = \frac{D^3 \rho W g}{6}$$

Again, my math could be wrong, but if this result is correct, it really challenges my intuition. Why isn't the torque on the dam just the average force due to water pressure times the "middle depth", D/2? The force varies linearly with depth, so why isn't the total torque just (average moment arm) x (average force) = FD/2?

4. Sep 3, 2015

TSny

That looks like the correct result, but it's a little confusing to use N for both the torque and the number of slabs.

Let Fi be the force on the ith slab and Li the moment arm of the ith slab. Let N be the number of slabs and assume N is an even number. Then, for large N the net torque is (approximately)

Torque = F1L1 + ⋅⋅⋅ + FN/2LN/2 +⋅⋅⋅ + FNLN .

Note that the one term FN/2LN/2 is essentially the average force times the average lever arm. So, it's clear that the total sum is greater than the average force times the average lever arm.

FN/2LN/2 is just the torque on the one slab in the middle.

Last edited: Sep 3, 2015